Indices, surds, quadratics, simultaneous equations, inequalities, polynomials, the factor theorem, partial fractions, graphs of functions, composite and inverse functions, the modulus function and graph transformations.

What this dot point is asking

AQA wants fluency with indices and surds, solving and analysing quadratics, simultaneous equations and inequalities, manipulating polynomials using the factor and remainder ideas, splitting expressions into partial fractions, working with composite, inverse and modulus functions, and applying the standard graph transformations. This dot point is the algebraic toolkit on which the whole of pure mathematics depends.

Indices and surds

The index laws are $a^m \times a^n = a^{m+n}$, $\dfrac{a^m}{a^n} = a^{m-n}$, $(a^m)^n = a^{mn}$, $a^0 = 1$, $a^{-n} = \dfrac{1}{a^n}$ and $a^{1/n} = \sqrt[n]{a}$. Fractional indices combine these, so $a^{m/n} = \sqrt[n]{a^m}$.

To rationalise a surd denominator, multiply numerator and denominator by the conjugate, which uses the difference of two squares to clear the surd. For example $\dfrac{1}{3 + \sqrt{2}} = \dfrac{3 - \sqrt{2}}{(3 + \sqrt{2})(3 - \sqrt{2})} = \dfrac{3 - \sqrt{2}}{9 - 2} = \dfrac{3 - \sqrt{2}}{7}$.

Quadratics and the discriminant

A quadratic $ax^2 + bx + c = 0$ has solutions $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Questions that say "find the values of $k$ for which..." almost always reduce to a discriminant condition: set $b^2 - 4ac$ equal to, greater than, or less than zero, and solve the resulting inequality or equation in $k$.

Simultaneous equations and inequalities

Solve a linear and a quadratic simultaneously by substitution, which usually yields a quadratic in one variable; the number of solutions is the number of intersection points. For inequalities, solve the corresponding equation first to find critical values, then test the sign of each interval (a sketch or sign table is the safest approach), remembering to reverse the inequality when multiplying by a negative.

Polynomials and the factor theorem

To factorise $f(x) = x^3 - 6x^2 + 11x - 6$, test small values: $f(1) = 1 - 6 + 11 - 6 = 0$, so $(x - 1)$ is a factor. Divide to obtain a quadratic factor, then factorise that, giving $(x - 1)(x - 2)(x - 3)$.

Partial fractions

A proper algebraic fraction with a factorised denominator can be split into simpler pieces, which is essential for later integration and binomial work.

Functions and transformations

A composite function $fg(x)$ means apply $g$ first then $f$. An inverse function $f^{-1}$ undoes $f$ and exists only when $f$ is one-to-one; its graph is the reflection of $y = f(x)$ in the line $y = x$, and its domain is the range of $f$. The modulus function $|x|$ gives the non-negative size of $x$, so $|{-3}| = 3$; solving modulus equations often needs both the positive and negative cases.